Let's analyze and solve the given problem step-by-step using the provided information: **System:** A + B ⇌ C + D **Feed:** A = 80 kmol/h B = 100 kmol/h **Relative volatilities:** α_A = 4 α_B = 2.5 α_C = 1.2 α_D = 1 **Equilibrium Constant:** K = 2e^(-334/T) **Activity coefficient:** γ_i = e^(2x_i²) **Reactor trays:** Number of trays = 10, 25, or 40 **Efficiency:** 60% ### 1. **Conversion** Let the conversion of A be \( X \). For the reaction A + B ⇌ C + D, let the initial moles (per hour) of A = 80 and B = 100. At conversion X, the moles of A reacted = 80X, So, at equilibrium: - A left = 80(1-X) - B left = 100 - 80X - C formed = 80X - D formed = 80X ### 2. **Product Purity** Product purity depends on how much C and D are produced and if there is any separation. ### 3. **Stage-wise Profiles** Not enough direct data is given; typically, you would use α and number of trays to perform a McCabe-Thiele type calculation, adjusting for reaction. ### 4. **Energy Integration** Not enough data specified for a diagram here. ### 5. **Effect of Decoupling Reaction** If reaction and separation are decoupled, conversion and separation may be less efficient than if they're integrated (reactive distillation). ### 6. **Comparison with Non-reactive Separation** Reactive separation allows shifting equilibrium, potentially achieving higher conversion and purity. --- #### **Key Calculation: Conversion Using Equilibrium Constant** Write the equilibrium expression: At equilibrium, \[ K = \frac{a_C \cdot a_D}{a_A \cdot a_B} \] Where \( a_i = \gamma_i x_i \) (activity coefficients and mole fractions). Let’s assume ideal solution, so \( \gamma_i ≈ 1 \), for a first approximation. Total moles at equilibrium = (80 - 80X) + (100 - 80X) + 80X + 80X = 180 Mole fractions: - \( x_A = \frac{80(1-X)}{180} \) - \( x_B = \frac{100-80X}{180} \) - \( x_C = \frac{80X}{180} \) - \( x_D = \frac{80X}{180} \) Using the equilibrium constant: \[ K = \frac{(x_C)(x_D)}{(x_A)(x_B)} \] \[ K = \frac{\left(\frac{80X}{180}\right)\left(\frac{80X}{180}\right)}{\left(\frac{80(1-X)}{180}\right)\left(\frac{100-80X}{180}\right)} \] \[ K = \frac{(80X)^2}{80(1-X)(100-80X)} \] \[ K = \frac{640X^2}{80 \cdot 180 (1-X)(100-80X)/180} \] \[ K = \frac{640X^2}{80(1-X)(100-80X)} \] Given \( K = 2e^{-334/T} \), you need T to get K. If T is not given, you can only express conversion in terms of K. So, solving: \[ K = \frac{640X^2}{80(1-X)(100-80X)} \] \[ K = \frac{80X^2}{(1-X)(100-80X)} \] \[ K(1-X)(100-80X) = 80X^2 \] You can solve this quadratic equation for X if K is given. --- ### **Final Answer (Summary)** The conversion at equilibrium, \( X \), is found from: \[ K(1-X)(100-80X) = 80X^2 \] where \( K = 2e^{-334/T} \) and the feed is A = 80 kmol/h, B = 100 kmol/h. For a specific temperature, plug in T to find K, then solve for X. The rest of the tasks (purity, stage profiles, etc.) depend on additional column design details and operational parameters, but the main calculation for conversion is summarized above.